Cho \(Q=\frac{5}{7}.\frac{13}{7^2}.\frac{97}{7^4}....\frac{3^{2^{99}}+2^{2^{99}}}{7^{2^{99}}}\)CMR: \(Q.\left(7^{2^{100}-1}\right)\)
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Cho \(Q=\frac{5}{7}.\frac{13}{7^2}.\frac{97}{7^4}.....\frac{3^{2^{99}}+2^{2^{99}}}{7^{2^{99}}}\)
Chứng minh rằng: \(Q.\left(7^{2^{100}-1}\right)\in N\).
Cho:
\(Q=\frac{5}{7}.\frac{13}{7^2}.\frac{97}{7^4}.....\frac{3^{2^{99}}+2^{2^{99}}}{7^{2^{99}}}\)
Chứng minh rằng: \(Q\left(7^{2^{100}-1}\right)\in N\).
Cho:
\(Q=\frac{5}{7}.\frac{13}{7^2}.\frac{97}{7^4}.....\frac{3^{2^{99}}+2^{2^{99}}}{7^{2^{99}}}\)
Chứng minh rằng: \(\left(Q.7^{2^{100}-1}\right)\in N\)
Cho:
\(Q=\frac{5}{7}.\frac{13}{7^2}.\frac{97}{7^4}.....\frac{3^{2^{99}}+2^{2^{99}}}{7^{2^{99}}}\)
Chứng minh rằng: \(Q\left(7^{2^{100}-1}\right)\in N\).
\(Tính.S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\)
\(CMR.\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
a,\(\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^3}\)
b,\(\left(-2\frac{1}{3}\right)^{100}.\left(-0,5\right)^{99}:\left(\frac{7}{3}\right)^{98}:\left(\frac{1}{4}\right)^{50}\)
mk doan la` de sai, sua: \(\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^3-2^{12}}\)
\(=\frac{3^7.\left(3^2-2^3\right)+2^{10}.\left(3^2-2^3\right)}{3^7.\left(3^3-2^2\right)+2^{10}.\left(3^3-2^2\right)}=\frac{3^7+2^{10}}{\left(3^7+2^{10}\right).24}=\frac{1}{24}\)
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Bài 1:a, Cho Sfrac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{9^2} .Chứng minh rằng frac{2}{5} S frac{8}{9}b, Tìm x thuộc z để phân số frac{x^2-5x-1}{x+2}có giá trị là số nguyênc, Chứng minh rằng left(frac{7}{65}+1right)left(frac{7}{84}+1right)left(frac{7}{105}+1right)left(frac{7}{124}+1right)...left(frac{7}{153+1}right)left(frac{7}{560}+1right) 2d, Chứng minh rằng frac{1}{3}-frac{2}{3^2}+frac{3}{3^3}-frac{4}{3^4}+frac{5}{3^5}-...+frac{99}{3^{99}}-frac{100}{3^{100}} frac{3}{16}
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Bài 1:
a, Cho S=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\) .Chứng minh rằng \(\frac{2}{5}< S< \frac{8}{9}\)
b, Tìm x thuộc z để phân số \(\frac{x^2-5x-1}{x+2}\)có giá trị là số nguyên
c, Chứng minh rằng \(\left(\frac{7}{65}+1\right)\left(\frac{7}{84}+1\right)\left(\frac{7}{105}+1\right)\left(\frac{7}{124}+1\right)...\left(\frac{7}{153+1}\right)\left(\frac{7}{560}+1\right)< 2\)
d, Chứng minh rằng \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\frac{5}{3^5}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
a, Tính : frac{left(13frac{1}{4}-2frac{5}{27}-10frac{5}{6}right).230frac{1}{25}+46frac{3}{4}}{left(1frac{3}{10}+frac{10}{3}right):left(12frac{1}{3}-14frac{2}{7}right)}b, Tính : Pfrac{frac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{2012}}{frac{2011}{1}+frac{2010}{2}+frac{2009}{3}+...+frac{1}{2011}}c, Tính : frac{left(1+2+3+...+99+100right)left(frac{1}{2}-frac{1}{3}-frac{1}{7}-frac{1}{9}right)left(63.1,2-21.3,6right)}{1-2+3-4+...+99-100}
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a, Tính : \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b, Tính : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c, Tính : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
Tìm x, biết:
a) \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)
b)\(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)
c) (x+2) - (x+3) >0
d)\(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)
a) \(\dfrac{x+5}{5}+\dfrac{x+5}{7}+\dfrac{x+5}{9}=\dfrac{x+5}{11}+\dfrac{x+5}{13}\)
\(\Rightarrow\left(x+5\right)\left(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\right)=\left(x+5\right)\left(\dfrac{1}{11}+\dfrac{1}{13}\right)\)
\(\Rightarrow\dfrac{143}{315}\left(x+5\right)=\dfrac{24}{143}\left(x+5\right)\)
\(\Rightarrow\dfrac{143}{315}\left(x+5\right)-\dfrac{24}{143}\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(\dfrac{143}{315}-\dfrac{24}{143}\right)=0\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
b) \(\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(3+\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=3+\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(1+\dfrac{x+2}{100}+1+\dfrac{x+3}{99}+1+\dfrac{x+4}{98}=1+\dfrac{x+5}{97}+1+\dfrac{x+6}{96}+1+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(\dfrac{100}{100}+\dfrac{x+2}{100}+\dfrac{99}{99}+\dfrac{x+3}{99}+\dfrac{98}{98}+\dfrac{x+4}{98}=\dfrac{97}{97}+\dfrac{x+5}{97}+\dfrac{96}{96}+\dfrac{x+6}{96}+\dfrac{95}{95}+\dfrac{x+7}{95}\)\(\Rightarrow\)\(\dfrac{x+102}{100}+\dfrac{x+102}{99}+\dfrac{x+102}{98}=\dfrac{x+102}{97}+\dfrac{x+102}{96}+\dfrac{x+102}{95}\)
\(\Rightarrow\)\(\left(x+102\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{98}\right)=\left(x+102\right)\left(\dfrac{1}{97}+\dfrac{1}{96}+\dfrac{1}{95}\right)\)
\(\Rightarrow\)\(x+102=0\)
\(\Rightarrow x=-102\)
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c) \(\left(x+2\right)-\left(x+3\right)>0\)
\(\Rightarrow x+2-x-3>0\Rightarrow-1>0\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)\left(x+\dfrac{7}{3}\right)\ge0\)
TH1: \(\left\{{}\begin{matrix}x-5\ge0\\x+\dfrac{7}{3}\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge5\\x\ge\dfrac{-7}{3}\end{matrix}\right.\)
\(\Rightarrow x\ge\dfrac{-7}{3}\)
TH2: \(\left\{{}\begin{matrix}x-5\le0\\x+\dfrac{7}{3}\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le5\\x\le\dfrac{-7}{3}\end{matrix}\right.\)
\(\Rightarrow x\le5\)
TH3: \(\left[{}\begin{matrix}x-5=0\\x+\dfrac{7}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)
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